Prove that the parallelogram circumscribing a circle is a rhombus.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Since ABCD is a parallelogram,
AB = CD ...(1)
BC = AD ...(2)
It can be observed that
DR = DS (Tangents on the circle from point D)
CR = CQ (Tangents on the circle from point C)
BP = BQ (Tangents on the circle from point B)
AP = AS (Tangents on the circle from point A)
Adding all these questions, we obtain
DR + CR +BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
On putting the values of equations (1) and (2) in this equation, we obtain
2AB = 2BC
AB = BC ...(3)
Comparing equations (1),(2) and (3), we obtain
AB = BC = CD = DAHence, ABCD is a rhombus.
Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at point P, Q, R, S. let us join the vertices of the quadrilateral ABCD to the center of the circle.
Consider ΔOAP and ΔOAS,
AP =AS (Trangents from the same point)
OP =OS (Radii of the same circle)
OA =OA (Common side)
ΔOAP≅ΔOAS (SSS congruence criterion)
And thus, ∠POA=∠AOS
Similarly, we can prove that ∠BOC+∠DOA=180∘
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.